JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Elasticity & SHM

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A 3 m long wire of radius 3 mm shows an extension of 0.1 mm when loaded vertically by a mass of 50 kg in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is

P \times 1 0^{11} Nm^{- 2}

, where the value of P is : (Take

g = 3 π m / s^{2}

)

(A)5✓ Correct
(B)10
(C)25
(D)2.5

Answer: (A)

Solution

Young's modulus is

Y = \frac{F L}{A Δ L}

. Here

F = m g = 50 \times 3 π N

L = 3 m

A = π r^{2} = π (3 \times 1 0^{- 3})^{2} m^{2}

and

Δ L = 0.1 mm = 1 0^{- 4} m

. Then

Y = \frac{150 π \times 3}{π \times 9 \times 1 0 ^{- 6} \times 1 0 ^{- 4}} = \frac{450}{9 \times 1 0 ^{- 10}} = 5 \times 1 0^{11} N m^{- 2}

. The value of

P

5

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option

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